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Copyright © 2022 by Ian Beardsley
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Contents
The Basis………………………………………………………………..3
1.0 Introduction………………………………………………………9
2.0 Giordano’s Relationship…………………………………….12
3.0 The Principle of Least Action……………………………..18
4.0 Proton-Seconds………………………………………………..20
5.0 The Natural Second…………………………………………..23
6.0 Rigorous Formulation of Proton Seconds……………27
7.0 The Theory……………………………………………………….31
8.0 Orbital Velocity………………………………………………..43
9.0 Logos………………………………………………………………51
10.0 Conclusion………………………………………………53
11.0 Summary…………………………………………………57
12.0 Graphic Equations………………………………………59
Matter can be explained as a construct of space and time at the fundamental level of elementary
particles, the elements of the periodic table follow from that construct. We demonstrate a
recurrent six-fold symmetry of Nature and formulate logos, her language in abstract terms.
of 3 67
The Basis The transformation that rotates counter-clockwise where
is given by the standard matrix
Equation 1
We suggest there is an aspect of Nature founded on six-fold symmetry, the example of
which we are interested in here is The Periodic Table of the Elements, because it has 18
groups which we can define by carbon, C. This because we have the following scenario:
Equations 2
And, we pull out the 2 and the 3 and write (Fig. 1)
, ,
Where , such that
which is given by
T : ℝ
2
→ ℝ
2
T(
x ) = A
x
A =
(
cos(θ) −sin(θ)
sin(θ) cosθ
)
3 + 3 + 3 = 9
2 + 2 + 2 = 6
3 ⋅ 6 = 18
2 ⋅ 9 = 18
2 ⋅ 3 = 6
2cos
π
4
= 2
2cos
π
5
=
5 + 1
2
2cos
π
6
= 3
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
Fig. 1 Dividing line in
golden mean,
of 4 67
In general
Equations 3 ,
, ,
And these can be mapped by the matrix A onto a linear vector
space (Fig. 2)
=
=
f (n) = 2cos
π
n
n = 4,5,6
π
4
= 45
∘
π /5 = 36
∘
π
6
= 30
∘
A =
(
2cos(θ ) −2sin(θ )
2sin(θ ) 2cosθ
)
A =
(
2cos(30
∘
) −2sin(30
∘
)
2sin(30
∘
) 2cos(30
∘
)
)
A
e
1
=
3 −1
1 3
(
1
0
)
= ( 3,1)
A
e
2
=
3 −1
1 3
(
0
1
)
= (−1, 3)
A =
(
2cos(36
∘
) −2sin(36
∘
)
2sin(36
∘
) 2cos(36
∘
)
)
Φ −
1
2
(5 − 5)
1
2
(5 − 5) Φ
Fig. 2!
of 5 67
=
Our is based on the square (Fig. 3)
is the line . The reflection through is given by:
Equation 9.4
And our is the equilateral triangle:
A
e
1
= Φ,
1
2
(5 − 5)
A
e
2
=
1
2
(5 + 5), Φ
A =
(
2cos(45
∘
) −2sin(45
∘
)
2sin(45
∘
) 2cos(45
∘
)
)
2 − 2
2 2
A
e
1
= 2, 2
A
e
2
= − 2, 2
2cos
π
4
π
4
= 45
∘
x
2
= x
1
x
2
= x
1
A =
(
0 1
1 0
)
2cos
π
6
Fig. 3
of 6 67
To transform the square into the equilateral triangle we expand the square of base
with the matrix
Equation 9.5
And we see becomes and we have added half the square to itself. (Fig. 4)
Or, better we can use the contraction
We draw in the diagonal of the the half-square and reassemble the two half-triangles
into an equilateral triangle (Fig. 8). To get we take the half square and draw in the
circle of radius 1/2. (Fig. 5) We have
e
1
A =
(
3/2 0
0 1
)(
1
0
)
=
3
2
,1
e
1
3/2
A =
(
1/2 0
0 1
)(
1
0
)
=
1
2
Φ
1
2
2
+ 1
2
=
1
4
+
4
4
=
5
2
5
2
+
1
2
=
5 + 1
2
= Φ
Fig. 4
of 7 67
Thus we see the periodic table is 18 groups (Fig. 6). Carbon is in group 14. We have
18-14=4 valence electrons. Hydrogen is neither a metal or a non-metal but ionizes like a
metal by losing one electron becoming and carbon being means it needs 4
positive ions to be neutral meaning it combines with 4 hydrogens to each C, or with two
hydrogens to a C and a C in long chains (hydrocarbons) which form the Skeltons of
organic compounds in life chemistry (Fig. 7) .
H
+
C
4−
Fig. 5
Fig. 6
Fig. 7
Fig. 8
of 8 67
That is:
is phenomenal because It allows multiplication between degrees and seconds to output
our fundamental ratios ( ). We see in the following wave:
, ,
Where t=1 second is carbon yielding:
And, t=6 seconds is hydrogen yielding:
1
6protons
⋅
1
α
2
m
p
h4π r
2
p
Gc
= 1.00second
2, 3, . . .
A = A
0
cos(θt)
A
0
= 1
θ = 30
∘
,60
∘
,45
∘
A(60
∘
) = cos(60
∘
⋅ 1s) = 0.5
A(30
∘
) = cos(30
∘
⋅ 1s) = 3/2
A(45
∘
) = cos(45
∘
⋅ 1s) = 2/2
A(60
∘
) = cos(60
∘
⋅ 6s) = 1
A(30
∘
) = cos(30
∘
⋅ 6s) = − 1
A(45
∘
) = cos(45
∘
⋅ 6s) = 0
of 9 67
1.0 Introduction Gravity is a property of space measured by the universal constant of gravity,
G:
Equation 1.1
Matter, or inertia, which measures matter’s ability to resist a force is for each particle (protons
and neutrons) we will suggest given by:
Equation 1.2.
Which describes mass per meter over time, which is:
Equation 1.3:
It must be adjusted by the fine structure constant . It is my guess the factor should be which
is 18,769.:
Equation 1.4.
Here we are suggesting that the proton and neutron are the 3-dimensional cross-sections of a
hypersphere. Thus we consider the surface area of a proton, :
Equation 1.5.
We take the square root to get meters:
Equation 1.6.
We multiply that with the value we have in equation 1.4:
Equation 1.7.
We find that the mass of a proton realizes if we divide this by six seconds:
Equation 1.8.
That is hydrogen. We see that the element carbon manifests if we divide instead by 1 second:
Equation 1.9.
Carbon (C) is the core element of life chemistry and it combines with hydrogen (H) to make the
skeletons of organic matter, the so-called hydrocarbons:
G = 6.674E − 11N
m
2
s
2
= 6.674E − 11
m
3
s
2
⋅ kg
h
Gc
=
kg ⋅ s
m
1.82E − 16
kg ⋅ s
m
α
1
α
2
(1.82E − 16kg ⋅ s /m)(18,769) = 3.416E − 12kg ⋅ s /m)
S
p
S
p
= 4π r
2
p
= 8.72E − 30m
2
S
p
= 2.953E − 15m
(2.953E − 15m)(3.416E − 12
kg ⋅ s
m
) = 1.009E − 26kg ⋅ s
m
p
m
p
=
1E − 26
6secon d s
= 1.67E − 27kg
m
c
=
1E − 26kg ⋅ s
1secon d
= 1E − 26kg = 6pr oton s = 6m
p
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Equation 1.10.
Equation 1.11.
If we divide 1E-26kgs by something greater than 6 seconds we get fractional protons. The rest of
the elements in the periodic table occur for dividing by something less that 1-second. It seems
the duration of a second is natural. If it is, since it was formed by a calendar based on
reconciling the periods of the moon and the sun in the earth sky, it should be in the Earth-moon
orbital mechanics. I find it is, that:
That is, the earth day (86,400 seconds) times the kinetic energy of the moon to the kinetic
energy of the earth is about 1 second (about 1.2 seconds).
All of the computations thus far are shown on the next page…
m
p
=
1E − 26kg ⋅ s
6secon d s
m
c
=
1E − 26kg ⋅ s
1secon d
K E
moon
K E
earth
(Ear th Da y) ≈ 1secon d
of 11 67
"
of 12 67
2.0 Giordano’s Relationship We can find our one second in terms of the atom that
predicts the mass of a proton. But I find we need to develop a constant I call k, which I
derived from Giordano’s relationship.
We now formulate what I call Giordano’s Relationship: Warren Giordano writes in his paper
The Fine Structure Constant And The Gravitational Constant: Keys To The Substance Of The
Fabric Of Space, March 21, 2019:
In 1980, the author had compiled a series of notes analyzing Einstein’s geometric to kinematic
equations, along with an observation that multiplying Planck’s constant ‘ ’ by ‘ ’, where ‘ ’
is the Fine Structure Constant, and multiplying by yielded Newton’s gravitational
constant numerically, but neglecting any units.
Let’s do that
(6.62607E-34Js)(1+1/137)(1E23)=6.6744E-11 Js
And it works, G is:
G=6.67408E-11 N(m2/kg2)
Let us reformulate this as:
Equation 2.1
Where and H=1 gram/atom
Because for hydrogen 1 proton is molar mass 1 gram, for carbon 6 protons is 6 grams and so on
for 6E23 atoms per gram. Thus,…
Since grams and atom cancel we can work in grams even though our equations are in kilograms.
Let us not write H, since formally it is grams per mole of hydrogen but write
We have:
Or,…
Equation 2.2
Where
h
1 + α
α
10
23
h
(1 + α)
G
N
A
H = 6.0003
kg
2
⋅ s
m
N
A
= Avaga dr o′ s − Num ber = 6.02E 23atom s /gra m
N
A
H = 6.02E 23
atom s
gram
⋅
1gram
atom
= 6.02E 23
ℍ = 1
gram
atom
h
(1 + α)
G
⋅ N
A
ℍ = 6.0003kg
2
⋅
s
m
h(1 + α)N
A
ℍ = 6Gx
of 13 67
Equation 2.3
Let us say we were to consider Any Element say carbon . Then in general
Equation 2.4
We have
and
Because there are six grams of protons in carbon which has 6 protons and 6 neutrons and a
molar mass of 12. We have
12-6=6 grams of protons in the 12 grams of protons and neutrons. Thus
And it follows that
We see in general since the atomic number Z is the number of protons in an atom that in general
this holds for all elements because
And,
Therefore we always have:
Equation 2.5
This works nicely because we formulated molar mass nicely; we said element one (hydrogen)
which is one proton and one electron has one gram for a mole of atoms. Historically this was
done because we chose carbon (element six) to have 12 grams per mole, and determined what
the mole was such that it would hold. The reason this works is that hydrogen is one proton and
has no neutrons, but carbon has twelve neutrons but since hydrogen doesn’t have any neutrons,
and the neutron has the same mass as the proton, and our theory makes use only of protons (in
this instance of its formulation) equation 2.3
x = 1.00kg
2
s
m
𝔼
ℂ
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
ℂ =
6gram s
6proton s
N
A
=
6(6E 23proton s)
6gram s
N
A
ℂ = 6E 23
h
(1 + α)
G
⋅ N
A
ℂ = 6.0003kg
2
⋅
s
m
𝔼
N
A
=
Z ⋅ 6E 23proton s
Z ⋅ gram s
𝔼 =
Z ⋅ gram s
Z ⋅ proton s
N
A
⋅ 𝔼 = 6E 23
x = 1.00kg
2
s
m
of 14 67
Comes out to have x equal to 1.00 (nearly) even. It is at this moment that we point out, because
it is important, that in equation 2.5
is not molar mass, and that is a variable determined by ; it is the number of a mole of
atoms multiplied by the number of protons in . The reason we point this out, though it may
already be clear, is we wish to find the physical theory behind it. That is we need to find the
physical explanation for equation 6.4
It is the integer 6 to 3 ten thousandths. Which classifies it as interesting because since it is in
kilograms, seconds, and meters, it may mean these units of measurement have some kind of a
meaning. We can in fact write it:
We know that
The fine structure constant squared is the ratio of the potential energy of an electron in the first
circular orbit to the energy given by the mass of an electron in the Bohr model times the speed of
light squared. To begin our search for the meaning of equation 1.4 we convert x, the factor of
1.00 to astronomical units, years, and solar masses, as these are connected to the orbit
of earth as it relates to the sun. We have:
=
We can now write
Eq 2.6.
This unit of AU/year is very interesting. It is not , which would be the Earth’s orbital
velocity, but is a velocity given by the earth orbital radius to its orbital period, which is quantum
mechanical in nature. It relates to earth as as a state, as we have with atoms, a number. We
N
A
⋅ 𝔼 = 6E 23
𝔼
N
A
𝔼
𝔼
h
(1 + α)
G
⋅ N
A
𝔼 = 6.0003kg
2
⋅
s
m
h
(1 + α)
G
⋅ N
A
𝔼 = 6.000kg
2
⋅
s
m
α
2
=
U
e
m
e
c
2
kg
2
⋅
s
m
kg
2
1
⋅
s
m
⋅
(1.98847E 30)
2
M
2
⊙
kg
2
1.4959787E11m
AU
⋅
year
3.154E 7s
1.8754341E64
M
2
⊙
⋅ year
AU
h
(1 + α)
G
⋅ N
A
𝔼
AU
year
= 8.2172E 32M
⊙
2π AU/year
of 15 67
multiply both sides by and we have earth velocity on the left and the units stay the same on
the right. But what we will do is return to the form in kg-m-s and leave it as an equation but put
in the Earth mean orbital velocity which is 29.79km/s (Zombeck, Martin V. 1982). We get:
Eq. 2.7
This brings up an interesting question: while we have masses characteristic of the microcosmos
like protons, and masses characteristic of the macrocosmos, like the minimum mass for a star to
become a neutron star as opposed to a white dwarf after she novas (The Chandrasekhar limit)
which is 1.44 solar masses, we do not have a characteristic mass of the intermediary world where
we exist, a truck weighs several tons and tennis ball maybe around a hundred grams. To find
that mass let us take the geometric mean between the mass of a proton and the mass of 1.44
solar masses. We could take the average, or the harmonic mean, but the geometric mean is the
squaring of the proportions, it is the side of a square with the area equal to the area of the
rectangle with these proportions as its sides. We have:
We multiply this by 1.44 to get 2.8634E30kg. The mass of a proton is .
We have the intermediary mass is:
Eq. 2.8
All we really need to do now is divide 2.7 by 2.8 and we get an even number that is the six of our
six-fold symmetry.
Eq. 2.9
The six of our six-fold symmetry.
We have something very interesting here. We have
This is:
Equation 2.10
Where k is a constant, given
Equation 2.11
We can take the velocity of earth as being 30,000 m/s by rounding it. We have
4π
2
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 422.787kg
M
⊙
= 1.98847E 30kg
m
p
= 1.67262E − 27kg
m
i
= (2.8634E 30)(1.67262E − 27) = 69.205kg
1
m
i
h
(1 + α)
G
⋅ N
A
𝔼 ⋅ v
e
= 6.1092 ≈ 6
1
69.205kg
6kg
2
s
m
v
e
= 6
k v
e
= 6
k =
1
800
s
m
30,000
800
= 37
1
2
of 16 67
Using , we write
37.5 = 6.123734357
1
6
⋅
1
α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
k v
e
= 6
1
α
2
m
p
h 4π r
2
p
Gck v
e
= 1proton ⋅ secon d
(K . E . Moon)(Ear th Da y)
(K . E . E ar th)
≈ 1secon d
1
α
2
m
p
h 4π r
2
p
Gck v
e
(K . E . E ar th)
(K . E . Moon)(Ear th Da y)
= 1proton
of 17 67
Eq. 2.12
The primordial element from which all others were made.
We can explicitly write the constant k:
Equation 6.13
It was the Indian physicist Chandrasekhar who found the limit in mass for which a white dwarf
will not have its gravity overcome the degeneracy pressure and collapse. The non-relativistic
equation is:
Equation 2.14
Let us approximate 0.77 with 3/4. Since we have our constant
And
Then
Equation 2.15
Since our constant k in terms the Chandrasekhar limit is
Equation 2.16
1
α
2
m
p
h 4π r
2
p
Gc
(K . E . E ar th)
(K . E . Moon)(Ear th Da y)
= 6proton s
1
α
2
m
p
h 4π r
2
p
Gck v
e
(K . E . E ar th)
(K . E . Moon)(Ear th Da y)
= ℍ
ℍ = hydrogen
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
k =
1
m
2
i
h
1 + α
G
⋅ N
A
𝔼
m
i
= Mm
p
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
ℏ = h /2 π
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
of 18 67
3.0 The Principle of Least Action Indeed if we are to find our unit of one second we need to
consider the Principle of Least Action. All things take the path of least action, and do this in the
least time. As such we can derive the law of the reflection of light that is the angle of incidence
equals the angle of reflection, or the law of refraction that light refracts in going from one
medium to another such that the sine of the angle of incidence equals the sine of the angle of
refraction. This requires that (see illustration next page):
Where c is the speed of light in air, say, and v its speed in water. If t is our six seconds and since
we have if because we are dealing with the cross-section of a
hypersphere:
The radius of a hydrogen atom is 120pm=1.2E-10m. Thus if we call v the velocity of a proton ,
then we have:
To travel the radius of a hydrogen atom. To travel half the radius of a hydrogen atom we we have
400,000 seconds. We have my constant k
Is one thousandth of the earth orbit. How long does it take light to travel this distance?
This is our one second in terms of the atom, but happens to be the time it takes light to travel
1/1000 diameters of the earth orbit. That is the diameter of the earth orbit is 1000 light-seconds.
t = r
p
(
c + v
c v
)
r
p
= 0.833E − 25m
c → c
v =
299,792,459m /s
2E 24
= 1.49896E − 16m /s ≈ = 1.5E − 16m /s
v
p
v
p
t = 1.2E − 10m
t =
1.2E − 10m
1.5E − 16m /s
= 800,000s
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼 = 800m /s
(400,000s)(800m /s) = 3.2E8m
1AU = 1.5E 8m
3.2E8m
1.5E11m /AU
= 2(0.001)AU
3.2E8m
299,792,459m /s
= 1secon d
of 19 67
of 20 67
4.0 Proton-Seconds We show carbon, the core element of life is six-fold symmetric
with hydrogen in terms of the natural constants that characterize space, time, and
matter:
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
We find one second gives six protons which is carbon:
Equation 4.1
We find six seconds gives 1 proton is hydrogen:
Equation 4.2
For time t greater than 6 seconds we have fractional protons. For t<6 we the have other
elements.
Is proton-seconds. Divide by time we have a number of protons because it is a mass
divided by the mass of a proton. But these masses can be considered to cancel and leave
pure number. We make a program that looks for close to whole number solutions so we
can create a table of values for problem solving.
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G : 6.67408 × 10
−11
N
m
2
kg
2
c : 299,792,459m /s
α : 1/137
1
α
2
m
p
h4π r
2
p
Gc
= 6proton ⋅ seconds = carbon(C )
1
α
2
m
p
h4π r
2
p
Gc
= 1proton ⋅ 6seconds = hydro gen(H )
1
α
2
m
p
h4π r
2
p
Gc
of 21 67
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?):
100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
A very interesting thing here is looking at the values generated by the program, the
smallest integer value 1 second produces 6 protons (carbon) and the largest integer
value 6 seconds produces one proton (hydrogen). Beyond six seconds you have
fractional protons, and the rest of the elements heavier than carbon are formed by
fractional seconds. These are the hydrocarbons the backbones of biological chemistry.
Here is the code for the program:
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792459,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than
100?): ");
scanf("%i", &n);
}
while (n>=101);
{
of 22 67
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment,
decpart);
}}}}
We need an interpretation of equation 2.1, which was
Matter is that which has inertia. This means it resists change in position with a force
applied to it. The more of it, the more it resists a force. We understand this from
experience, but what is matter that it has inertia?
In this analogy we are suggesting a proton is a three dimensional bubble embedded in a
two dimensional plane. As such there has to be a normal vector holding the higher
dimensional sphere in a lower dimensional space. Thus if we apply a force to to the
cross-section of the sphere in the plane there should be a force countering it
proportional to the normal holding it in a lower dimensional universe. It is actually a 4-
dimensional hypersphere whose cross-section is a sphere. This counter force would be
experienced as inertia. (Fig. 1.1)
Since plank’s constant h is a measure of energy over time where space and time are
concerned it must play a role. Of course the radius of a proton plays a role since squared
and multiplied by it is the surface area of our proton embedded in space. The
gravitational constant is force produced per kilogram over a distance, thus it is a
measure of how the surrounding space has an effect on the proton giving it inertia. The
speed of light c has to play a role because it is the velocity at which events are separated
through time. The mass of a proton has to play a role because it is a measurement of
inertia itself. And alas the fine structure constant describes the degree to which these
factors have an effect. We see the inertia then in equation 6 is six protons over 1 second,
by dimensional analysis.
1
α
2
m
p
h4π r
2
p
Gc
= 6proton ⋅ seconds = carbon(C )
4π
of 23 67
We see the radius of a proton is given by carbon (1 second):
Equation 4.3.
The experimental radius of a proton is:
Equation 4.4.
The fine structure constant squared is the ratio of the potential energy of an electron in
the first circular orbit to the energy given by the mass of an electron in the Bohr model
times the speed of light squared:
Equation 4.5.
5.0 The Natural Second In that we get one second for carbon and 6 seconds for
hydrogen very nearly even, that is
Eq. 5.1
It is suggested that the second is a natural unit. If it is, since it comes from designing a
calendar that reconciles the phases of the moon with the Earth year (12 moons per year,
approximately) it is suggested the unit of a second should be in the Earth-Moon-Sun
orbital mechanics. The translational kinetic energy of the moon and earth are:
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
r
p
= 0.833
±
0.014 f m
α
2
=
U
e
m
e
c
2
1
6
⋅
1
α
2
m
p
h4π r
2
p
Gc
= 1.004996352seconds
K . E . Moon = 3.67E 28J
K . E . Ear th = 2.649E 33j
Fig 1.1
of 24 67
To find the translational kinetic energy of the moon:
Distance from earth: 3.85E8m
Orbital period:
T=27.32 days=2.36E6seconds
v=1.025E3m/s~1000m/s
Mass: 7.34767E22kg
Use
E=3.67E28 Joules
To find the translational kinetic energy of the earth:
Distance from Sun: 1AU=1.496E11m
Orbital period: 1 year=3.1558E7 seconds
v=2.9785E4m/s, Earth mass: 5.9722E24kg
E=2.649E33 Joules
Earth day=(24)(60)(60)=86,400 seconds.
It turns out:
Eq . 5.2
Where the Lunar Month can be as much as 31 days and is based on the lunar orbital
period (27.32 days). We have
Eq. 5.3
Essentially we have formed a Planck constant, h, for the moon by multiplying its kinetic
energy over the time for the period of its orbit:
2π (3.85E8m) = 2.419E 9m
E =
1
2
mv
2
2π (1.496E11m) = 9.399E11m
1second ≈
(KEof Moon)(Lun arOrbitalPeriod )
Luna rMonth
EarthDay
(KEof Ear th)
31days
(1Ear th Day)
= 31 ≈ π
3
= 31.006
h = (3.67E 28J )(2.36E6s) = 8.6612E 34J ⋅ s
8.6612E 34J ⋅ s
KEof Ear th
= 32.696seconds
of 25 67
If we let the lunar month cancel with moon’s orbital period, we have:
Eq. 5.4
Since
is units of mass divided by we can let it cancel with , the mass of a proton, and
write:
Eq. 5.5
That is:
is phenomenal because It allows multiplication between degrees and seconds to output
our fundamental ratios ( ). We see in the following wave:
, ,
Where t=1 second is carbon yielding:
And, t=6 seconds is hydrogen yielding:
1
α
2
m
p
h4π r
2
p
Gc
≈ 6protons ⋅
KEof Moon
KEof Ear th
⋅ Ear th Day
1
1sec
⋅
1
α
2
m
p
h4π r
2
p
Gc
m
p
m
p
1
α
2
m
p
h4π r
2
p
Gc
≈ 6 ⋅
KEof Moon
KEof Ear th
⋅ Ear th Day = (6)1.2secon ds
1
6protons
⋅
1
α
2
m
p
h4π r
2
p
Gc
= 1.00second
2, 3, . . .
A = A
0
cos(θt)
A
0
= 1
θ = 30
∘
,60
∘
,45
∘
A(60
∘
) = cos(60
∘
⋅ 1s) = 0.5
A(30
∘
) = cos(30
∘
⋅ 1s) = 3/2
A(45
∘
) = cos(45
∘
⋅ 1s) = 2/2
A(60
∘
) = cos(60
∘
⋅ 6s) = 1
A(30
∘
) = cos(30
∘
⋅ 6s) = − 1
A(45
∘
) = cos(45
∘
⋅ 6s) = 0
of 26 67
In so far as
relates carbon=1second to the Earth-Moon-Sun orbital mechanics and to the radius of a
proton through six-fold symmetry:
We have the following around which all our structure is based
,
, ,
1
α
2
m
p
h4π r
2
p
Gc
≈ 6 ⋅
KEof Moon
KEof Ear th
⋅ Ear th Day ≈ (6)1secon d
1
α
2
m
p
h4π r
2
p
Gc
= 6
f (n) = 2cos
π
n
n = 2,3,4,5.6.,...
f (2) = 2
f (5) = Φ = 1/ϕ
f (6) = 3
of 27 67
6.0 Rigorous Formulation of Proton Seconds
We can actually formulate this differently than we have. We had
But if t1 is not necessarily 1 second, and t6 is not necessarily six seconds, but rather t1 and t2 are
lower and upper limits in an integral, then we have:
Equation 6.1
This Equation is the generalized equation we can use for solving problems.
Essentially we can rigorously formulate the notion of proton-seconds by considering
Equation 6.2
Is protons-seconds squared where current density is and ( can also be
). We say
Equation 6.3
Keeping in mind q is not charge (coulombs) but a number of charges times seconds, here a
number of protons. It is
Equation 6.4
Dividing Equation 6.2 through by by t:
Equation 6.5
Which is proton-seconds. Dividing through by t again:
1
t
1
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton s
1
t
6
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
2
t
1
1
t
2
dt = ℕ
∫
t
qdt = t
2
∬
S
ρ(x, y, z) d x d y
J = ρ
v
ρ = Q /m
3
ρ
Q /m
2
Q =
∫
V
ρd V
ℕ =
1
α
2
m
p
h 4π r
2
p
Gc
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
= t
∬
S
ρ(x, y, z) d x d y
of 28 67
Equation 6.6
We see that if where and then J is I/m2 (current per square meter)
is analogous to amperes per per square meter which are coulombs per second through a surface.
Thus we are looking at a number of protons per second through a surface. Thus we write:
Is carbon where 0.5 seconds is magnesium (Mg) from the values of time corresponding to
protons in the output from our program and 1.0 seconds is carbon (C). We see we have the
following theorem:
Equation 6.7
So as an example,…
Is fluorine (F). Divide by xy with x=y=1 and we have current density. And multiply by 1 second
which is carbon and we have protons per square meter.
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
2
= proton s
J = ρ
v
ρ = Q /m
3
v = m /s
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
C
t
Mg
dt
t
2
= 6
∫
1.0
0.5
t
−2
dt = − 6(1 − 2) = 6
1
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
3
=
∬
S
J ⋅ d
S
1
α
2
m
p
h 4π r
2
p
Gc
∫
1.0
0.5
dt
t
3
=
∬
S
J ⋅ d
S = − 3
(
1 −
1
0.25
)
= 9
proton s
secon d
J(x, y, z) = (0,0, J ) = − J
k
d
S = d xd y
k
J ⋅ d
S = (0,0, J ) ⋅ (0,0, d xd y) = − Jd xd y
of 29 67
We are now equip to do computations in proton-seconds. We use equation 7.6 from two to
three, the smallest prime numbers that multiply to make six-fold symmetry in our hexagonal
proton that we found described its radius (My feeling is we introduce the factor of 2 because
carbon is 6 protons +6 neutrons and 2 times 6 is twelve):
But we already know this is the radius of a proton in fm, it is as if fm is a natural length, as well
as the second we have been using. 2 seconds in our program is 3 protons (Lithium Li) and 3
seconds is two protons is helium, He:
By what value would you like to increment?: 0.01
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
86.1425 protons 0.070000 seconds 0.142517 decpart
50.2498 protons 0.120000 seconds 0.249805 decpart
43.0713 protons 0.140000 seconds 0.071259 decpart
40.1998 protons 0.150000 seconds 0.199841 decpart
30.1499 protons 0.200000 seconds 0.149876 decpart
26.2173 protons 0.230000 seconds 0.217283 decpart
25.1249 protons 0.240000 seconds 0.124893 decpart
24.1199 protons 0.250000 seconds 0.119900 decpart
23.1922 protons 0.260000 seconds 0.192213 decpart
20.0999 protons 0.300000 seconds 0.099920 decpart
17.2285 protons 0.350000 seconds 0.228504 decpart
15.0749 protons 0.400000 seconds 0.074944 decpart
14.0232 protons 0.430000 seconds 0.023204 decpart
13.1086 protons 0.460000 seconds 0.108647 decpart
12.0600 protons 0.500000 seconds 0.059957 decpart
11.1666 protons 0.540000 seconds 0.166627 decpart
2
α
2
m
p
h 4π r
2
p
Gc
∫
t
dt
t
2
= proton s
2
α
2
m
p
h 4π r
2
p
Gc
∫
3
2
dt
t
3
= 0.833proton s /secon d
of 30 67
10.2203 protons 0.590000 seconds 0.220302 decpart
10.0500 protons 0.600000 seconds 0.049964 decpart
9.1363 protons 0.660000 seconds 0.136332 decpart
8.1486 protons 0.740000 seconds 0.148621 decpart
8.0400 protons 0.750000 seconds 0.039971 decpart
7.1785 protons 0.839999 seconds 0.178546 decpart
7.0941 protons 0.849999 seconds 0.094093 decpart
7.0116 protons 0.859999 seconds 0.011603 decpart
6.2165 protons 0.969999 seconds 0.216474 decpart
6.1530 protons 0.979999 seconds 0.153040 decpart
6.0909 protons 0.989999 seconds 0.090889 decpart
By what value would you like to increment?: 0.25
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
24.1199 protons 0.250000 seconds 0.119904 decpart
12.0600 protons 0.500000 seconds 0.059952 decpart
8.0400 protons 0.750000 seconds 0.039968 decpart
6.0300 protons 1.000000 seconds 0.029976 decpart
4.0200 protons 1.500000 seconds 0.019984 decpart
3.0150 protons 2.000000 seconds 0.014988 decpart
2.1927 protons 2.750000 seconds 0.192718 decpart
2.0100 protons 3.000000 seconds 0.009992 decpart
1.2060 protons 5.000000 seconds 0.205995 decpart
1.1486 protons 5.250000 seconds 0.148567 decpart
1.0964 protons 5.500000 seconds 0.096359 decpart
1.0487 protons 5.750000 seconds 0.048691 decpart
1.0050 protons 6.000000 seconds 0.004996 decpart
0.2487 protons 24.250000 seconds 0.248659 decpart
0.2461 protons 24.500000 seconds 0.246121 decpart
0.2436 protons 24.750000 seconds 0.243635 decpart
Thus we write
Equation 6.8.
Thus we suggest the radius of a proton of a proton (0.833protons) is the proton radius .
Now we integrate from Aluminum is 13 protons=0.372 seconds to phosphorus is 15 protons =
0.396 seconds which is to integrate across silicon:
By what value would you like to increment?: 0.006
How many values would you like to calculate for t in equation 1 (no more than 100?): 100
251.2490 protons 0.024000 seconds 0.248978 decpart
59.1174 protons 0.102000 seconds 0.117416 decpart
50.2498 protons 0.120000 seconds 0.249805 decpart
40.1998 protons 0.150000 seconds 0.199844 decpart
37.2221 protons 0.162000 seconds 0.222080 decpart
27.1621 protons 0.222000 seconds 0.162058 decpart
25.1249 protons 0.240000 seconds 0.124907 decpart
20.0999 protons 0.300000 seconds 0.099916 decpart
17.0338 protons 0.354000 seconds 0.033823 decpart
16.2096 protons 0.372000 seconds 0.209604 decpart
15.2272 protons 0.396000 seconds 0.227202 decpart
2
α
2
m
p
h 4π r
2
p
Gc
∫
t
He
t
Li
dt
t
3
= 0.833proton s /secon d = r
proton
r
proton
of 31 67
14.1549 protons 0.426000 seconds 0.154862 decpart
13.2236 protons 0.456000 seconds 0.223620 decpart
13.0519 protons 0.462000 seconds 0.051885 decpart
12.1084 protons 0.498000 seconds 0.108375 decpart
11.1666 protons 0.540000 seconds 0.166615 decpart
11.0439 protons 0.546000 seconds 0.043906 decpart
10.1515 protons 0.594000 seconds 0.151471 decpart
Equation 6.9.
Silicon can be doped with phosphorus to make negative (n-type) silicon that semi-conducts thus
enabling the construction of logic circuits that you can use to make computing machines. But
this must be joined with positive (p-type) silicon which usually uses boron, but boron is in the
same group as aluminum, just above it. This results in a theory for AI elements as mathematical
constructs, that we will go into now.
7.0 The Theory
Above we see the artificial intelligence (AI) elements pulled out of the periodic table of the
elements. As you see we can make a 3 by 3 matrix of them and an AI periodic table. Silicon and
germanium are in group 14 meaning they have 4 valence electrons and want 4 for more to attain
noble gas electron configuration. If we dope Si with B from group 13 it gets three of the four
electrons and thus has a deficiency becoming positive type silicon and thus conducts. If we dope
the Si with P from group 15 it has an extra electron and thus conducts as well. If we join the two
types of silicon we have a semiconductor for making diodes and transistors from which we can
make logic circuits for AI.
As you can see doping agents As and Ga are on either side of Ge, and doping agent P is to the
right of Si but doping agent B is not directly to the left, aluminum Al is. This becomes important.
I call (As-Ga) the differential across Ge, and (P-Al) the differential across Si and call Al a dummy
in the differential because boron B is actually used to make positive type silicon.
That the AI elements make a three by three matrix they can be organized with the letter E with
subscripts that tell what element it is and it properties, I have done this:
2
α
2
m
p
h4π r
2
p
Gc
∫
t
Al
t
P
dt
t
3
= − 6(6.377 − 7.22) = 5protons/second = boron
of 32 67
Thus E24 is in the second row and has 4 valence electrons making it silicon (Si), E14 is in the
first row and has 4 valence electrons making it carbon (C). I believe that the AI elements can be
organized in a 3 by 3 matrix makes them pivotal to structure in the Universe because we live in
three dimensional space so the mechanics of the realm we experience are described by such a
matrix, for example the cross product. Hence this paper where I show AI and biological life are
mathematical constructs and described in terms of one another.
We see, if we include the two biological elements in the matrix (E14) and and (E15) which are
carbon and nitrogen respectively, there is every reason to proceed with this paper if the idea is to
show not only are the AI elements and biological elements mathematical constructs, they are
described in terms of one another. We see this because the first row is ( B, C, N) and these
happen to be the only elements that are not core AI elements in the matrix, except boron (B)
which is out of place, and aluminum (Al) as we will see if a dummy representative makes for a
mathematical construct, the harmonic mean. Which means we have proved our case because the
first row if we take the cross product between the second and third rows are, its respective unit
vectors for the components meaning they describe them!
The Computation
E
13
E
14
E
15
E
23
E
24
E
25
E
33
E
34
E
35
A = (Al, Si, P)
B = (G a, Ge, As)
A ×
B =
B
C
N
Al Si P
Ga Ge As
= (Si ⋅ As − P ⋅ G e)
B + (P ⋅ G a − Al ⋅ A s)
C + (Al ⋅ G e − Si ⋅ G a)
N
A ×
B = − 145
B + 138
C + 1.3924
N
A = 26.98
2
+ 28.09
2
+ 30.97
2
= 50g/mol
B = 69.72
2
+ 72.64
2
+ 74.92
2
= 126g/mol
A ⋅
B = A Bcosθ
cosθ =
6241
6300
= 0.99
θ = 8
∘
A ×
B = A Bsinθ = (50)(126)sin8
∘
= 877.79
of 33 67
And silicon (Si) is at the center of our AI periodic table of the elements. We see the biological
elements C and N being the unit vectors are multiplied by the AI elements, meaning they
describe them. But we have to ask; Why does the first row have boron in it which is not a core
biological element, but is a core AI element? The answer is that boron is the one AI element that
is out of place, that is, aluminum is in its place. But we see this has a dynamic function.
The Dynamic Function
The primary elements of artificial intelligence (AI) used to make diodes and transistors, silicon
(Si) and germanium (Ge) doped with boron (B) and phosphorus (P) or gallium (Ga) and arsenic
(As) have an asymmetry due to boron. Silicon and germanium are in group 14 like carbon (C)
and as such have 4 valence electrons. Thus to have positive type silicon and germanium, they
need doping agents from group 13 (three valence electrons) like boron and gallium, and to have
negative type silicon and germanium they need doping agents from group 15 like phosphorus
and arsenic. But where gallium and arsenic are in the same period as germanium, boron is in a
different period than silicon (period 2) while phosphorus is not (period 3). Thus aluminum (Al)
is in boron’s place. This results in an interesting equation.
Equation 7.1.
The differential across germanium crossed with silicon plus the differential across silicon
crossed with germanium normalized by the product between silicon and germanium is equal to
the boron divided by the average between the germanium and the silicon. The equation has
nearly 100% accuracy (note: using an older value for Ge here, it is now 72.64 but that makes the
equation have a higher accuracy):
Due to an asymmetry in the periodic table of the elements due to boron we have the
harmonic mean between the semiconductor elements (by molar mass):
Equation 7.2
This is Stokes Theorem if we approximate the harmonic mean with the arithmetic mean:
877.79 = 29.6g/m ol ≈ Si = 28.09g/m ol
Si(A s − G a) + Ge(P − Al )
SiGe
=
2B
Ge + Si
28.09(74.92 − 69.72) + 72.61(30.97 − 26.98)
(28.09)(72.61)
=
2(10.81)
(72.61 + 28.09)
0.213658912 = 0.21469712
0.213658912
0.21469712
= 0.995
Si
B
(A s − G a) +
Ge
B
(P − Al ) =
2SiGe
Si + G e
∫
S
( ∇ × u ) ⋅ d S =
∮
C
u ⋅ d r
of 34 67
We can make this into two integrals:
Equation 7.3.
Equation 7.4
If in the equation (The accurate harmonic mean form):
Equation 7.5
We make the approximation
Equation 7.6
Then the Stokes form of the equation becomes
Equation 7.7
Thus we see for this approximation there are two integrals as well:
Equation 7.8
Equation 7.9
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
d xd y ≈
1
Ge − Si
∫
Ge
Si
xd x
∫
1
0
∫
1
0
Si
B
(As − Ga)dyd z ≈
1
3
1
(Ge − Si)
∫
Ge
Si
xd x
∫
1
0
∫
1
0
Ge
B
(P − Al )d xdz ≈
2
3
1
(Ge − Si)
∫
Ge
Si
yd y
Si
B
(As − Ga) +
Ge
B
(P − Al ) =
Ge − Si
Ge
Si
dx
x
2SiGe
Si + Ge
≈ Ge − Si
∫
1
0
∫
1
0
[
Si
B
(As − Ga) +
Ge
B
(P − Al )
]
dyd z =
∫
Ge
Si
d x
∫
1
0
∫
1
0
Si
B
(As − Ga)dyd z =
1
3
∫
Ge
Si
dz
∫
1
0
∫
1
0
Ge
B
(P − Al )d ydz =
2
3
∫
Ge
Si
dz
of 35 67
For which the respective paths are
One of the double integrals on the left is evaluated in moles per grams, the other grams
per mole (0 to 1 moles per gram and 0 to 1 grams per mole).
By making the approximation
In
We have
Equation 7.10
is the differential across Si, is the differential across Ge
and is the vertical differential.
Which is Ampere’s Circuit Law
We see if written
Which is interesting because it is semiconductor elements by molar mass which are used
to make circuits.
We say (Phi) is given by
y
1
=
1
3
B
SiGa
ln(z)
y
2
=
2
3
B
Si Al
ln(z)
2SiGe
Si + Ge
≈ Ge − Si
Si( As − Ga)
B
+
Ge(P − Al )
B
=
2SiGe
Si + Ge
Si
ΔGe
ΔS
+ Ge
ΔSi
ΔS
= B
ΔSi = P − Al
ΔGe = As − Ga
ΔS = Ge − Si
∇ ×
B = μ
0
J + μ
0
ϵ
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
Φ
of 36 67
and
And
=1.618
=0.618
(phi) the golden ratio conjugate. We also find
Equation 7.11
Thus since
And we have
Equation 7.12
We see and are both and c is in the Si (silicon) field wave, but for E and B fields c is the
speed of light.
a = b + c
a
b
=
b
c
Φ = a /b
ϕ = b/a
ϕ
(ϕ)ΔGe + (Φ)ΔSi = B
∇ ×
B = μ
J + μϵ
0
∂
E
∂t
Si
ΔGe
ΔS
= B − Ge
ΔSi
ΔS
ΔG e =
ΔS
Si
B −
G e
Si
ΔSi
(
∇
2
−
1
c
2
∂
2
∂t
)
E = 0
(
∇
2
−
1
c
2
∂
2
∂t
)
B = 0
c =
1
ϵ
0
μ
≈ ϕ
μ
ϵ
0
Φ
ϕ
ϵ
0
= 8.854E − 12F ⋅ m
−1
μ = 1.256E − 6H /m
G e
Si
= μ ϵ
0
of 37 67
To find the Si wave our differentials are
It is amazing how accurately we can fit these differentials with an exponential equation for the
upward increase. The equation is
This is the halfwave:
ΔS
Si
= μ
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
Si = 0
(
∇
2
−
1
ϕ
2
∂
2
∂x
)
G e = 0
ΔC = N − B = 14.01 − 10.81 = 3.2
ΔSi = P − Al = 30.97 − 26.98 = 3.99
ΔGe = As − Ga = 74.92 − 69.72 = 5.2
ΔSn = Sb − In = 121.75 − 114.82 = 6.93
ΔPb = Bi − T l = 208.98 − 204.38 = 4.6
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
of 38 67
Equation 7.13
Building Another Matrix
We pull these AI elements out of the periodic table of the elements to make an AI periodic table:
y(x) = e
0.4x
+ 1.7
y(x) = e
2
5
x
+
17
10
y(x) = e
B
Al
x
+
A g
Cu
B
Al
=
10.81
26.98
= 0.400667
A g
Cu
=
107.87
63.55
= 1.6974 ≈ 1.7
of 39 67
We now notice we can make a 3 by 3 matrix of it, which lends itself to to the curl of a vector field
by including biological elements carbon C (above Si):
=
=
Which results in Stokes theorem:
Where
We then able to write this with product notation
While we have the AI BioMatrix
i
j
k
∂
∂x
∂
∂ y
∂
∂z
( − C ⋅ P) y (Si ⋅ G a)z (Ge ⋅ As)y
(Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
[
(72.64)(74.92) − (28.09)(69.72)
]
i +
[
(12.01)(30.97)
]
k
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a = ex p
(
1
Ge − Si
∫
Ge
Si
ln(x)d x
)
∇ ×
u = (Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
d
a =
(
zd yd z
i + yd yd z
k
)
u = ( − C ⋅ P)y
i + (Si ⋅ G e)z
j + (G a ⋅ As)y
k
5
∫
Ge
Si
∫
Ge
Si
∇ ×
u ⋅ d
a =
n
n
∏
i=1
x
i
of 40 67
Which we use to formulate a similar equation. We can form another 3X3 matrix we will call the
electronics matrix:
We can remove the 5th root sign in the above equation by noticing
=(28.085)(72.64)(12.085)(107.8682)(196.9657)=
Where we have substituted carbon (C=12.01) the core biological element for copper (Cu).
But since we have:
We take the ratio and have
Almost exactly 3 which is the ratio of the perimeter of regular hexagon to its diameter used to
estimate pi in ancient times by inscribing it in a circle:
5
∏
i=1
x
i
= Si ⋅ G e ⋅ Cu ⋅ Ag ⋅ Au
523,818, 646.5
g
5
m ol
5
∫
Ge
Si
∫
Ge
Si
( ∇ ×
u) ⋅ d
a = 170,535, 359.662(g/mol )
5
523,818, 646.5
170,535, 359.662
= 3.0716
of 41 67
Perimeter=6
Diameter=2
6/2=3
Thus we have the following equation…
Showing The Calculation using the most accurate data possible…
Ge=72.64
As=74.9216
Si=28.085
Ga=69.723
C=12.011
P=30.97376200
=
=
=154,082,837.980+16,452,521.6822=
π = 3.141 . . .
π
∫
Ge
Si
∫
Ge
Si
( ∇ ×
u) ⋅ d
a =
5
∏
i=1
x
i
(Ge ⋅ As − Si ⋅ Ga)
i + (C ⋅ P)
k
[
(72.64)(74.9216) − (28.085)(69.723)
]
i +
[
(12.011)(30.97376200)
]
k
3,484 . 134569
(
g
m ol
)
2
i + 372.025855
(
g
m ol
)
2
k
∫
Ge
Si
∫
Ge
Si
(
3,484 . 134569
(
g
m ol
)
2
i + 372.025855
(
g
m ol
)
2
k
)
⋅
(
zd yd z
i + yd yd z
k
)
∫
Ge
Si
∫
Ge
Si
(
3,484 . 134569
(
g
m ol
)
2
⋅ zd zd y + 372.025855
(
g
m ol
)
2
⋅ yd zd y
)
∫
Ge
Si
3,484 . 134569
(
(72.64 − 28.085)
2
2
)
d y +
∫
Ge
Si
372.025855y ⋅ (72.64 − 28.085)d y
3458261.42924
(
g
m ol
)
4
(72.64 − 28.085) + 16575.6119695
(
g
m ol
)
3
(
(72.64 − 28.085)
2
2
)
of 42 67
=
(28.085)(72.64)(12.085)(107.8682)(196.9657)=
Where we have substituted carbon C=12.01 for copper Cu. We use Cu, Ag, Au because they are
the middle column of our electronics matrix, they are the finest conductors used for electrical
wire. We use C, Si, Ge because they are the middle column of our AI Biomatrix. Si and Ge are the
primary semiconductor elements used in transistor technology (Artificial Intelligence) and C is
the core element of biological life. We have
Perimeter/Diameter of regular hexagon = 3.00
The same value as our 3.0716 if taken at two places after the decimal.
170,535, 359.662
(
g
m ol
)
5
5
∏
i=1
x
i
= Si ⋅ G e ⋅ C ⋅ A g ⋅ Au
523,818, 646.5
g
5
m ol
5
523,818, 646.5
170,535, 359.662
= 3.0716
π = 3.141 . . .
3.141 + 3.00
2
= 3.0705
of 43 67
8.0 Orbital Velocity Earth rotates through 15 degrees in one hour:
The distance then that the earth equatorial surface rotates through is where r is the earth
radius and theta is in radians.
Now we look at how many degrees through which the earth rotates in 1 minute:
For seconds…
Now let’s look at the same for the distance the earth moves around the Sun in an hour, a minute,
and a second as well…
And finally…
360
24
= 15
s = r θ
s = (6,378.14)(0.2618ra d ) = 1669.8k m /hr
24(60) = 1440min
360
1440
= 0.25
∘
= 0.0043633r a d
s = (6,378.14)(0.00436) = 27.83k m /min
24(60)(60) = 86,400s
360
∘
86,400
= 0.004167
∘
= 0.00007272r a d
s = (6,378,14)(0.00007272) = 0.464k m /s
(365.25)(24) = 8766hr /yr
360
8766
= 0.041
∘
= 0.00071678r a d
(149,598,000)(0.000716768) = 107,227k m /hr
(365.25)(24)(60) = 525960
360
525960
= = 0.000684463
∘
= 0.000011946r a d
s = (149,598,000)(0.000011946) = 1,787.1k m /min
(365.25)(24)(60)(60) = 31557600
of 44 67
We have the following table:
As we can see I am in good agreement with Martin Zombek, Handbook of Space Astronomy and
Astrophysics which provides data. Notice 27.83 km/min in earth rotation is approximately the
29.786 km/sec in earth orbit. That is a valuable clue. Now let us consider
Earth: 1 AU=149,598,000km (average earth-sun separation).
The earth rotates through about one degree a day.
Earth Orbit:
Earth Rotation: r=6,378.14km
Moon: sidereal month =27.83 days, r=405,400 km
360
31557600
= 0.000011408
∘
= 0.000000199r a d
(149,598,000)(0.000000199) = 29.786k m /sec
360
365.25
= 0.9856
deg
d a y
≈ 1
deg
d a y
360
∘
= 2π ra di a n s = 6.283
ra d
year
0.9856
∘
= 0.017202424
ra d
d a y
(149,598,000)(0.017202424) = 2,573,448.201
k m
d a y
360
∘
1d a y
= 6.283
ra d
d a y
s = r θ = (6,378.14)(6.283) = 40,075.0
k m
d a y
of 45 67
The points to be made in this exploration
1. So the earth goes through 1 degree a day and the moon 1 kilometer per second.
2. Earth rotates through 27.83 km/min at its equatorial surface and orbits through 29.786km/s
around the sun. These two are close to the same.
3. The synodic month is 29.53 days approximately equals the 29.786 km/s the Earth moves
around the sun. The sidereal month is 27.83 days is the 27.83 km/min through which the
Earth rotates at its surface.
With that information we have this mystery of sexagesimal in the earth-moon-sun orbital
parameters, solved. We see we can make the following equation in table form:
Consider minute times day:
Thus,..
The circle is divided into 360 units, each unit (each degree) is the distance the earth moves
around the Sun in a day, where a day is one turn of the earth on its its axis, and as such there are
360 such turns in the time it takes the earth to go around the sun approximately (365.25 days).
We have:
1 astronomical unit (AU) is the distance of the earth from the sun on the average, and is always
close to that because its orbit is approximately circular. We have
(27.83)(24) = 667.92
h ours
m onth
360
667.92
= 0.529
∘
= 0.0094
ra d
hr
(405,400k m)(0.0094ra d /hr) = 3,810.76k m /hr
(3,810.75k m /hr)(hr /60min) = 63.5k m /min = 1.0585k m /s
orbit rotat ion orbit m oon
29.786k m minute 1d a y k ilom eter
secon d 27.83k m degree secon d
=
min ⋅ d a y
deg
⋅
k m
s
2
(min)(d a y) = 60(24 ⋅ 60 ⋅ 60) = 864,000sec
2
min ⋅ d a y
deg
⋅
k m
s
2
=
86,400s
2
deg
⋅
k m
s
2
= 86,400 ⋅
k m
deg
86,400 ⋅
k m
deg
⋅ 360
∘
= 311,040,000k m
311,040,000k m
149,598,000k m /AU
= 2.079AU
of 46 67
This is approximately the diameter of the Earth orbit. We define our variables:
Earth orbits:
Earth rotates:
Earth orbits:
Moon orbits:
Earth completes a 360 degree orbit yields:
Where on the right it is in radians and is the radius of the Earth’s orbit. We have
This is (0.00618)360=2.225
0.00618 is
Where is the inverse of the golden ratio. .
We have:
Equation 8.1
Our base ten counting is defined
is defined
, such that
which is given by
v
e
= 29.786
k m
s
ω
e
=
27.83k m
min
=
27.83k m
min
⋅
min
60sec
= 0.4638
k m
sec
θ
e
=
1d a y
deg
=
(24)(60)(60)
deg
= 14,400
sec
deg
v
m
=
1k m
sec
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅
360
∘
1AU
= 2π
v
e
⋅ θ
e
⋅ v
m
ω
e
⋅ r
e
r
e
= 1AU
(29.786)(14,400)(1k m /s)
(0.4638)(149,598,000)
360 = 2.225AU
ϕ
100
ϕ
ϕ = 0.618 =
1
Φ
ϕ
100
360
∘
= 2AU
10
0
= 1,10
1
= 1,10
2
= 100,...
ϕ
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
of 47 67
Thus since the diameter of the Earth orbit is
Then its radius is
Since we measure time with the Earth orbital period and that period is given by Kepler as
Then
Equation 8.2
Is approximately one year. In this section we set out two show the historical development of the
second by dividing up the motions of the Earth, Moon, and the apparent motion of the Sun into
units of 24, and 60 result in the solar system’s size as based around the Earth orbit giving us
Where this duration of a second is in the atomic world in carbon (six protons) the core element
of life which we found was:
And that the duration of a second was as we showed in the orbital mechanics of the Earth and
moon as such
Since we have established a connection between the microcosmos and the macrocosmos we
would do well to introduce the units of AU (astronomical unit), year, solar masses. Thus we want
to know the universal gravitational constant in these units:
ϕ
100
360
∘
= 2AU
1
2
⋅
ϕ
100
360
∘
= 1AU
T
2
= a
3
T =
3
2
10 ⋅
ϕ
3
2
10
3
⋅ 360
∘
ϕ
100
360
∘
= 2AU
1
6proton s
⋅
1
α
2
m
p
h 4π r
2
p
Gc
= 1.00secon d
(K . E . Moon)
(K . E . E ar th)
(Ear th Da y) ≈ 1secon d
of 48 67
For a year we have
(365.25)(24)(60)(60)=31557600 seconds
And for an AU
1 AU=1.496E11m
We can immediately put this to work. In order for the earth to stay in orbit its centripetal force
must equal the gravitational force. Its orbital velocity must be given by:
Since in our theory we present R as:
Then
Equation 8.3
Which evaluates:
G = 6.67408E − 11
m
3
kg ⋅ s
2
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅
AU
3
3.3479E 33m
3
⋅
1.98847E 30kg
M
⊙
⋅
9.9477E14s
2
year
2
G = 39.433
AU
3
M
⊙
⋅ year
2
G ≈ 40
AU
3
M
⊙
⋅ year
2
GM
⊙
= 40
AU
3
year
2
GM
⊙
4π
2
= 0.99885 ≈ 1
G
Mm
R
2
=
mv
2
r
v =
GM
R
1
2
⋅
ϕ
100
360
∘
= 1AU
v =
5
3
GM
ϕ
of 49 67
Converting this to meters per second:
Which should be about right. The orbital velocity is given in the data tables as about 30,000m/s
is an average over a varying velocity due to the Earth’s slightly elliptical orbit.
Since we have
For the earth, we show now it is true for the moon as well, and indeed it would hold for any
system circular, or approximately circular. For the Sun and its planets
If you use astronomical units and Earth years. In general we use
Which has the constant of proportionality
If m is small compared with M, you can write
Since we want to work with Earth and the Moon m is not so small compared to M. So,…
=
=
And these do round neatly to one-exponent like this. The sidereal lunar month is
(27.83)(24)(60)(60)=2,404,513 seconds. Thus,…
v =
5
3
(40)(1)
0.618
= 6
AU
year
6
AU
year
⋅
year
3.156E 7
⋅
1.496E11m
AU
= 28,441
m
s
1
2
⋅
ϕ
100
360
∘
= 1AU
T
2
= a
3
T
2
= G
M + m
4π
2
a
3
1
k
= G
M + m
4π
2
1
k
= G
M
4π
2
1
k
= (6.674E − 11)
(
5.9722E 24kg + 7.347731E 22kg
4π
2
)
(6.674E − 11)
(
6E 24kg
4π
2
)
4E14
4π
2
=
1E14
π
2
= 1E13
of 50 67
The tables give
Which is an average for the lunar orbit in its slightly elliptical orbit.
We need to find G in terms of Lunar Units (LU) as opposed to Astronomical Units (AU), and in
terms of the lunar month as one instead of the earth year. We see G has the same value we had
for solar-earth units:
=
We only need to convert to m, kg, s by using their relationships with LU, LM, and . That is,
for the orbital velocity is always six:
Where is the orbital velocity of the moon, and is the orbital velocity of the earth. The
difference comes when converting the system from lunar units in the case of the first to kg/m/s
and from astronomical units in the case of the second to kg/m/s. For example the first is:
Which is correct, the tables give that the lunar orbital velocity is 1000 m/s. This factor of six is
the factor recurrent throughout our work in this paper, like
a
3
= (1E13)(2,404,512)
2
= 5.78E 25
a = 3.8668E8m
a = 3.833E8m
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
LU
3
(3.8668E8m)
3
⋅
5.9722E 24kg
M
⊕
⋅
(2,404,512s)
2
L M
2
40.896 ⋅
LU
3
M
⊕
⋅ L M
2
M
⊕
v
m
=
5
3
(40)(1)
ϕ
= 6
v
e
=
5
3
(40)(1)
ϕ
= 6
v
m
v
e
v
m
= 6
3.8668E8m
2,404,512s
= 964.886m /s
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
of 51 67
9.0 Logos
To find logos, is to find Nature not just as number, but as dimensionless whole
number, because fractions have inherent in them irrational numbers, which are
unending decimal expressions that at some point must be rounded to so many
figures to put them to use. Logos then is the language of Nature that transcends
conundrum because it would be utilized as relationship between point, plane, and
line as opposed to units. We have found that for the orbital velocity:
Equation 9.1
Where is the earth orbital radius. This results in the orbital velocity for a orbiting body is
Equation 9.2.
Which is logos because it is number, which is 6. The orbital velocity of a body is always 6
because:
Where is the orbital velocity of the moon, and is the orbital velocity of the earth, That is G
is always 40, and M is always 1 providing the orbit is circular. Let us show this for Venus. Its
orbital distance VU (Venus units) is 1.082E11meters. It orbital period is 1.94E7s is the Venus
year (VY).
=
Thus we have
We can then express all orbital velocities as 6, but to find there values in a formal system of
units, we need to convert from these natural units to something like kg/m/s. Thus logos
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
e
v =
5
3
GM
ϕ
v
m
=
5
3
(40)(1)
ϕ
= 6
v
e
=
5
3
(40)(1)
ϕ
= 6
v
m
v
e
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
LU
3
(1.082E11m)
3
⋅
1.989E 30kg
M
⊙
⋅
(1.94E 7s)
2
V Y
2
39.44
V U
3
M
⊙
V Y
2
≈ 40
v
v
=
5
3
(40)(1)
ϕ
= 6
of 52 67
translated into a language we understand can best be done in a square array, or as a matrix
transformation. We have
You will find this gives , ,
All of which are correct within the variations of these velocities in their deviations from a
perfectly circular orbit. Logos is:
Equations 9.3
, such that
which is given by
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
v
m
= 964.886m /s
v
e
= 28,443m /s
v
v
= 33,463.9m /s
(
5
3
GM
ϕ
5
3
GM
ϕ
5
3
GM
ϕ
)
R
m
T
m
R
e
T
e
R
v
T
v
=
v
m
v
e
v
v
=
6
6
6
R
m
=
1
2
⋅
ϕ
100
360
∘
= 1LU
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
v
=
1
2
⋅
ϕ
100
360
∘
= 1V U
v =
5
3
GM
ϕ
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
of 53 67
10.0 Conclusion To say that logos is:
Is to say that for the solar system Kepler’s Law of Planetary Motion holds:
That in general
Because as such for Earth we have
For the Moon we have
For Venus we have
Making
, ,
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
T
2
= a
3
T
2
= G
M + m
4π
2
a
3
G = 6.67408E − 11
m
3
kg ⋅ s
2
⋅
AU
3
3.3479E 33m
3
⋅
1.98847E 30kg
M
⊙
⋅
9.9477E14s
2
year
2
G ≈ 40
AU
3
M
⊙
⋅ year
2
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
LU
3
(3.8668E8m)
3
⋅
5.9722E 24kg
M
⊕
⋅
(2,404,512s)
2
L M
2
40.896 ⋅
LU
3
M
⊕
⋅ L M
2
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
V U
3
(1.082E11m)
3
⋅
1.989E 30kg
M
⊙
⋅
(1.94E 7s)
2
V Y
2
39.44
V U
3
M
⊙
V Y
2
≈ 40
v
e
=
5
3
(40)(1)
ϕ
= 6
v
m
=
5
3
(40)(1)
ϕ
= 6
v
v
=
5
3
(40)(1)
ϕ
= 6
of 54 67
This is derived from
Which is to say
Is logos because
such that
which is given by
Which is the relationship between point, plane, and line.
G
Mm
R
2
=
mv
2
R
v =
GM
R
1
2
⋅
ϕ
100
360
∘
= 1AU
R
m
=
1
2
⋅
ϕ
100
360
∘
= 1LU
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
R
v
=
1
2
⋅
ϕ
100
360
∘
= 1V U
v =
5
3
GM
ϕ
5 + 1
2
= Φ =
a
b
a
b
=
b
c
a = b + c
ϕ =
b
a
=
1
Φ
a
2
b
2
−
a
b
− 1 = 0
of 55 67
From everything we have said G, the gravitational constant is about 40
We can do this for any planet and get G is approximately 40. We found the orbital velocity of any
planets is 6. This is true because as we have shown, the orbital distance of any planet is
This gives since
That orbital velocities are for the moon, earth, and venus
But what does this mean? It means since
And, v=6, that where the orbital velocities of the planets are 6, their distances from the sun are
all one.
This gives:
G = 6.674E − 11
m
3
kg ⋅ s
2
⋅
AU
3
(Ear th Di sta n ceMeters)
3
⋅
Ear th MassKilogra ms
M
⊕
⋅
(Ear thOrbita l Per io d Secon d s)
2
L M
2
1
2
⋅
ϕ
100
360
∘
= 1AU
v =
GM
R
=
5
3
GM
ϕ
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
R =
GM
v
2
R =
40
36
≈ 1
(
1 1 1
)
3.8668E8m
1.496E11m
1.082E11m
=
R
m
R
e
R
v
of 56 67
But where six is
Six is also
And one is
And one is also
Where one second is given by Earth-moon orbital mechanics
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
1
1secon d
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton s = carbon(C )
(
1 1 1
)
3.8668E8m
1.496E11m
1.082E11m
=
R
m
R
e
R
v
1
6secon d s
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton = hydr ogen(H )
(K . E . Moon)
(K . E . E ar th)
(Ear th Da y) ≈ 1secon d
of 57 67
11.0 Summary What we have done here is capture the six-fold logos across the whole
spectrum:
Which are the biological life skeletons the hydrocarbons. Shown that carbon the core element of
life is one second predicts the radius of the proton:
That the second itself
Is a natural unit
We have brought this together:
Have described all orbital velocities as 6:
Have brought in chemistries’ Avogadro’s number with a constant k that that determines the
integer 6 in terms of the earth orbital velocity above:
,
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon d s = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon d = hydrogen(H )
r
p
=
18
3
α
2
m
p
Gc
4πh
= 8.288587 × 10
−16
m = 0.829f m
1
6
⋅
1
α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon d s
(K . E . Moon)
(K . E . E ar th)
(Ear th Da y) ≈ 1secon d
1
α
2
m
p
h 4π r
2
p
Gc
≈ 6proton s ⋅
K Eof Moon
K Eof Ear th
⋅ Ear th Day
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼
k v
e
= 6
of 58 67
But where six is
Six is also
And one is
And one is also
Where one second is given by Earth-moon orbital mechanics
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
1
1secon d
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton s = carbon(C )
(
1 1 1
)
3.8668E8m
1.496E11m
1.082E11m
=
R
m
R
e
R
v
1
6secon d s
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton = hydr ogen(H )
(K . E . Moon)
(K . E . E ar th)
(Ear th Da y) ≈ 1secon d
of 59 67
12.0 Graphic Equations We can say visually since the orbital velocity of a planet or a moon is
6, that it is the perimeter of a unit regular hexagon in logos:
Since the radius R of the orbit in logos is 1, then its visual logos is
the radius R of this regular hexagon because the radius of a
regular hexagon is equal to its side (s). We have
, ,
The six equilateral triangles that make up the regular hexagon are
computed here:
The height of each equilateral triangle is . Since then and since
is a regular hexagon, is a square because :
And, is a regular pentagon which contains is the golden ratio is :
R =
1
2
⋅
ϕ
100
360
∘
v =
5
3
GM
ϕ
G = 40
3
2
3
2
= cos
π
6
3 = 2cos
π
6
π
6
π
4
π
4
= 45
∘
π
5
Φ
1
ϕ
of 60 67
These are the regular polygons that tessellate (tile a plane without leaving gaps). They tessellate
in the plane, except for the regular pentagon, which tessellates closed, it encloses a volume, and
is 1 of 4 other solids that do the same which are made up of our tessellating regular polygons, the
equilateral triangle, the regular hexagon, and the square:
The area of a regular pentagon is given by
For s=1 this is . This gives the surface are of a regular dodecagon is:
A =
1
4
5(5 + 2 5 s
2
A = 1.72 ≈ 1.732 = 3
of 61 67
Actually let us denote the surface area of the dodecagon as
And reserve A for the area of a regular pentagon, which we
can write either
Or,…
Much in the same way that in vector calculus
Are both surface integrals. We want to use surface area of a pentagon as
Even though it shows a dodecahedron, because it speaks of the connection of the regular
pentagon to the dodecahedron. We want denote surface areas and volumes
V
= Volume of Dodecahedron
∫
S
d xd y =
∬
S
d xd y
of 62 67
With actual images and shapes of these objects, because logos is to have a deeper understanding
of Nature, and this allows us to work with the ideas in our mind in connection to the equations.
We have seen that the orbit of any planet or moon can be taken as 1:
And that 1 is astronomical unit (AU), moon unit (MU), Venus unit (VU), Mars unit (MU),…and
so on. The gravitational constant G can be taken as 40 for every planet, but in units that
correspond to the planet (or body) in particular:
Where M is the mass of the body orbited, like the Sun for the Earth, the Earth for the moon, and
is the planet’s orbital radius, it’s orbital period and these in any system of units, for
example in kg/m/s for the Earth
For the moon is the orbital radius in meters from the Earth, is the Earth mass, the mass
of the body it orbits, is its orbital period. The orbital velocity of any planet can be taken as 6
And must be translated from this the logos to a system of units we understand, like kg/m/s. In
general the components of logos are given by
n=1,2,3,4,5,6,…
In that
, , where , ,…
We have orbital velocity is hexaperimeter, and orbital radius is hexaradius:
R
p
=
1
2
⋅
ϕ
100
360
∘
= 1
G = 6.674E − 11
m
3
kg
2
⋅ s
2
⋅
M
R
p
T
p
R
p
T
p
R
e
= 1.496E1m = 1AU = Ear thOrbit
M
⊙
= 1.98847E 30kg = SolarMa ss
T
e
= 31,557,600s = Ear thYear
R
m
M
⊕
T
m
v =
5
3
GM
ϕ
=
5
3
(40)(1)
0.618
= 6
f (n) = 2cos
π
n
2cos
π
6
= 3
2cos
π
5
= Φ
ϕ =
1
Φ
2cos
π
4
= 2
of 63 67
We are going to suggest that where orbital velocity is associated with the regular hexagon, the
orbital radius is associated with the square, and the centripetal acceleration is associated with
the dodecahedron and pentagon. We have the mass of the earth in solar masses is
Which is correct. Where
=
Thus we have for the orbit radii, orbital velocities, and centripetal accelerations of an orbiting
body
M
⊕
=
5.9722E 24
1.98847E 30
= 3.00E − 6M
⊙
= 0.00003M
⊙
a
e
=
v
2
e
R
e
=
6
2
1
= 36
AU
year
2
R
e
=
1
2
⋅
ϕ
100
360
∘
= 1AU
2cos(36
∘
) = 2cos
360
∘
10
=
1
2ϕ
360
∘
10
⋅
1
cos
−1
1
2ϕ
= 1
360
∘
10
= cos
−
1
1
2ϕ
= 36
∘
24 cos(36
∘
) ≈ 20
(
1 1 1
)
3.8668E8m
1.496E11m
1.082E11m
=
R
m
R
e
R
v
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
(
6
2
6
2
6
2
)
3.8668E8m
(2,404,512s)
2
1.496E11m
(31,557,600s)
2
1.082E11m
(19,400,000s)
2
=
a
m
a
e
a
v
of 64 67
Thus we say
That,…
Is orbital radius, is:
Is orbital velocity, is:
Is centripetal acceleration, is:
Where we have used
The image of an icosahedron as the area of a regular hexagon, because a
regular hexagon is its shadow.
(
1 1 1
)
3.8668E8m
1.496E11m
1.082E11m
=
R
m
R
e
R
v
(
6 6 6
)
3.8668E8m
2,404,512s
1.496E11m
31,557,600s
1.082E11m
19,400,000s
=
v
m
v
e
v
v
(
6
2
6
2
6
2
)
3.8668E8m
(2,404,512s)
2
1.496E11m
(31,557,600s)
2
1.082E11m
(19,400,000s)
2
=
a
m
a
e
a
v
of 65 67
Aside from our logos formulation of the planets being useful theoretically, it is useful for orbital
mechanics computations whether it be of planets or spacecraft because it uses multiplication
of integer values like six for orbital velocity that can be converted to units we understand in
matrix form, which is at the heart of writing computer programs in their linear algebra
applications which to the computer are arrays. Further since we translate from logos to units,
by dividing out large numbers (like astronomical units and seconds in a year) in such a way that
they are not overly large numbers, yet large enough that they can be written as integers without
losing much accuracy and there are many simple programs on the internet that multiply
matrices with integer elements. This one comes from programmingsimplified.com:!
#include <stdio.h>
int main(int argc, const char * argv[]) {
int m,n,p,q,c,d,k,sum=0;
int first[10][10],second[10][10],multiply[10][10];
printf("Enter number of rows and columns of matrix 1: \n");
scanf("%d %d",&m,&n);
printf("Enter elements of matrix 1: \n");
for(c=0;c<m;c++)
for(d=0;d<n;d++)
scanf("%d",&first[c][d]);
printf("Enter number of rows and column of matrix 2: \n");
scanf("%d %d", &p,&q);
if(n!=p)
printf("The multiplication can't be done.");
else
{
printf("Enter the elements of matrix 2: \n");
for(c=0;c<p;c++)
for(d=0;d<q;d++)
scanf("%d", &second[c][d]);
for(c=0;c<m;c++){
for(d=0;d<q;d++){
for(k=0;k<p;k++){
sum=sum+first[c][k]*second[k][d];
}
multiply[c][d]=sum;
sum=0;
}
}
printf("Product of the matrices is: \n");
for(c=0;c<m;c++){
for(d=0;d<q;d++)
printf("%d\t", multiply[c][d]);
printf("\n");
}
}
return 0;
}
of 66 67
We need to divide out our data for the planets and the moon, which are 9 elements, which works
great because our program works only up to 3 by 3 matrices. We have
{3.8668E8m}{2,404,512s^2}=161=moon
{1.496E11m}{31,557,600s^2}=4,740=earth
{1.082E11m}{19,400,000s^2}=5,577=venus
You can do the rest of the planets, but we will just do these and put zeros in for the remaining 6
elements. Running the program we have
Enter number of rows and columns of matrix 1:
3
3
Enter elements of matrix 1:
161
0
0
4740
0
0
5577
0
0
Enter number of rows and column of matrix 2:
3
3
Enter the elements of matrix 2:
6
0
0
6
0
0
6
0
0
Product of the matrices is:
966 0 0
28440 0 0
33462 0 0
These answers are in meters per second.These answers are fairly accurate but
can be made more accurate by adjust the orbital velocity 6 according to
orbital pecularities, which would be a subject in itself that is outside the
scope of our work here.
of 67 67
The Author
